`x/a + y/b = 2`

`ax + by = a^2 - b^2`

` `Multiply first equation by `a^2.`

`a^2 (x/a + y/b = 2) = ax + a^2/by = 2a^2`

Subtract 2nd equation from 1st to get:

`a^2/by - by = a^2+b^2`

Factor y from left side to get:

`y(a^2/b - b) = a^2+b^2`

Simplify parentheses to be:

`y ((a^2-b^2)/b) = a^2 + b^2`

` `Multiply both sides by `b/(a^2 - b^2)`

**This yields** `y = (b (a^2+b^2))/(a^2 - b^2)`

First equation states that `x/a + y/b = 2,`

so **solving for x**: `x/a = 2 - y/b`

`x = a(2 - y/b)`

Substitute solution for y: `x = a(2 - 1/b( (b(a^2+b^2))/(a^2 - b^2))`

**The b's will cancel and distribute a, we get:**

`a((2a^2-2b^2 - a^2-b^2) / (a^2 - b^2))`

`a (( a^2 - 3b^2) / (a^2 - b^2))`

`x = (a^3 - 3ab^2) / (a^2 - b^2)`

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x/a +y/b =2

a*x +b*y =a^2+b^2

we multiply first eq, by a^2

a*x +(a^2/b)*y =2*a^2

a*x +b*y =a^2-b^2

we substract the second eq from the first one

y*(a^2/b -b)= 2a^2-a^2+b^2

y*(a^2-b^2)/b = a^2+b^2

y = b*(a^2+b^2)/(a^2-b^2)

From first eq we have

x/a =2-y/b

x/a =[(2a^2-2b^2)-(a^2+b^2)]/(a^2-b^2)

x =a*(a^2-3b^2)/(a^2-b^2)